The Parabola (Edexcel International A Level (IAL) Further Maths: Further Pure 1): Revision Note

You are given the Cartesian and parametric equations of a parabola in the Formulae Booklet.

A parabola has the parametric equations $x=a{t}^2$ and $y=2at$ where $a>0$.

Show that its Cartesian equation is $y^2=4ax$.

To find the Cartesian equation, eliminate $t$ from the parametric equations
It is easier to make $t$ the subject of $y=2at$

$t=\frac{y}{2a}$

Substitute this into the equation for $x$ and simplify

$\begin{array}{rcl}x& =& a{\left(\frac{y}{2a}\right)}^2\\ x& =& \frac{a{y}^2}{4a^2}\\ x& =& \frac{y^2}{4a}\end{array}$

Make $y^2$ the subject

$y^2=4ax$

You are given the focus and directrix of a parabola in the Formulae Booklet, but not the focus-directrix property.

The diagram shows the point $S\left(a,0\right)$, the vertical line $x=-a$ and a general point $P\left(x,y\right)$ that can move in the plane.

If $P$ is restricted to always be the same distance from $S$ as it is horizontally from the vertical line, use coordinate geometry to prove that it must lie on the parabola $y^2=4ax$.

It helps to add the point X on to the diagram, on the line $x=-a$  horizontally from P
The question says that PS  = PX
The distance PS  can be found using Pythagoras' theorem 

$P{S}^2={\left(x-a\right)}^2+y^2$

The distance PX  can be found from the diagram

$PX=x+a$

Square PS  = PX  and substitute in the results above

$\begin{array}{rcl}P{S}^2& =& P{X}^2\\ {\left(x-a\right)}^2+y^2& =& {\left(x+a\right)}^2\end{array}$

Expand and simplify both sides

$\begin{array}{rcl}{x}^2-2ax+a^2+y^2& =& x^2+2ax+a^2\\ -2ax+y^2& =& 2ax\\ y^2& =& 4ax\end{array}$

$P\left(x,y\right)$ must lie on this curve

$y^2=4ax$